let listNodes = [ 1,2,3,4,5,6,7,8,9,10 ]
        function reverseKGroup(listNode,k){
            if (k<1 || k > listNode){
            return listNode
            }
            let resultList = []
            let otherN = listNode.length%k
            // console.log(otherN)
            let times = (listNode.length - otherN)/k
            // console.log(times)
            for (let i = 0; i < times; i++) {
            let subList = []
            // 分组 k倒序遍历
            for (let j = k-1; j >= 0; j--) {
                let node = listNode[i*k+j]
                subList.push(node)
            }
            resultList.push(...subList)
            }
            // 把不满k的剩余链表拼接上
            if (otherN > 0){
            let otherList = listNode.slice(-otherN,listNode.length)
            resultList = resultList.concat(otherList)
            }
            return resultList
        }
        let reversNode = reverseKGroup(listNodes,6)
        console.log(reversNode)